GEOMETRY Chapter One—Answerstestbanktop.com/wp-content/uploads/2016/12/Downloadable... · 2016-12-31 · GEOMETRY Chapter Two—Answers True–False. 1. F:The triangles may be different

GEOMETRY Chapter One—Answers

True–False.1. T2. F—A better choice might be Use a Variable.3. T4. T5. F—The Look for a Pattern strategy is probably best.6. F—Solve a Simpler Problem is indicated.7. T8. F: It is “Look for a Pattern.”9. T

10. F: It is “Make a Table.”11. F—The fourth term might be 11 (Sequence of primes.).12. T13. T

Multiple Choice.14. d 15. a 16. e 17. e 18. d19. e 20. a 21. c 22. e 23. a24. e

Fill in the Blanks.25. Draw a Picture26. Make a Table27. Solve a Simpler Problem28. 180�29. Look for a Pattern30. deductive31. inductive

Writing.32. 1. Understand the Problem

2. Devise a Plan3. Carry Out the Plan4. Look Back

33. Grandmother, Mary, Aunt Suzie, Father, and Richard34. 15135. Inductive reasoning proceeds from the specific to the general. Deductive reasoning begins by

assuming general statements from which other specific statements are inferred.36. Answers will vary; here is one possibility

Figure Number 1 2 3 4 5 n

Number of dots 1 7 7 � 4 7 � 4 � 4 Should be: 2 less than the figure for 7� 4 � 4 � 4 the number of inner heptagons.(check with Each inner heptagon adds 4 new dots.

diagram) The nth heptagonal figure has 7 � (4 � (n � 2)) dots.

The method was to look at the components of the problem to “Look for a Pattern” and thedata was organized using the “Make a Table” strategy.

37. 27

Exercises/Problems.38. 339. 540. 84�41. Martha42. There are 16 � 9 � 4 � 1 � 30 squares.

43.

44. 16545.

46. 6th term � 35; 10th term � 99; nth term � n2 � 147. 36548. 21/6449.

Dimensions 1 � 1 � 24 1 � 2 � 12 1 � 3 � 8 1 � 4 � 6 2 � 2 � 6 2 � 3 � 4

Sides exposed 2 � (1 � 24 2 � (2 � 12 2 � (3 � 8 2 � (4 � 6 2 � (4 � 2 � (6 � 12 � 24) � 98 � 24) � 56 � 24) � 70 � 24) � 68 12 � 12) � 56 � 12) � 60

The minimum number of sides exposed is 56 and the maximum number of sides exposed is 98.

50. (a) 1, 4, 5, 9,(b) 2, , 6, , 16,(c) 3, , 27 (since; 3 � x � x � 3 � x � 27; 3x � 21, x � 7)

51. (a) 7th term � , 12th term � , nth term � , each term goes up by � 3.

(b) 7th term � , 12th term � , nth term � .

(c) 7th term � , 12th term � , nth term �12

3n12

3127

729

(n2 + 3n + 4)12

16229

3n36187, 10, 17

2610414, 23, 37

16 1 � 1 squares 9 2 � 2 squares

4 3 � 3 squares 1 4 � 4 squares

90 Chapter 1–Answers

Applications.52. 1853. Eleven of the 15-foot lengths and nine of the 8 foot lengths54. 655. 22.5�, 67.5�, 90�56. 457. 12058. 25� by 20�59. 67.5 minutes60. 1440�61. 862. There is more than one solution; but the carrier must start at N or M (an odd vertex) and stop

at N or M. The carrier cannot start and stop at the same house.

63. (a) (5 � 5) � (4 � 5) � (3 � 5) � (2 � 5) � (1 � 5) � 75(b) (n � n) � ((n � 1) � n) � . . . � (1 � n)

64. Here are two possible solutions

65. There are 24 cubes

Description None 8 corners, 4 top edges and 4 bottom edges (16)

None 4 centers 4 sides

Faces Painted 4 – 6 3 2 1 0

SIDE

CENTER

BOTTOM EDGE

TOP EDGE

CORNER

1

0

3

3

3

a. b.

Sample Solution: NAKJLIOMBNHKLEFOGHIEDCM A 1 9

1011

12

13 19 15

14

2021

22

2

3

4 5 6 7

817

18 16

K

J

N B

M

CD

E F

OIL

GH

Chapter 1–Answers 91

66. There are 20 cubes

67. The square arrangement encloses the most area.Top and

bottom rows 10 paversSides

(no corners) 2 pavers

9 pavers

3 pavers

8 pavers

4 pavers

7 pavers

5 pavers

Area 16 ft2 21 ft2 24 ft2 25 ft2

Description None 4 top edges, 4 bottom edges and 4 centers (12)

8 corners None

Faces Painted 5 – 6

4CENTER

4 BOTTOM EDGE

4TOP EDGE

CORNER3

4 3 0–2


GEOMETRY Chapter Two—Answers

True–False.1. F: The triangles may be different sizes and and may be different lengths.2. F: An angle of degree measure 180� is a straight angle while a reflex angle has measure

180� � n � 360�.3. T4. F—If the points are collinear they are contained in infinitely many planes.5. T6. F: Two obtuse angles would sum to greater than 180�.7. T8. F: Only regular triangles, squares and regular hexagons.9. F—The parallel sides are called bases.

10. T11. F: It is semiregular.12. F: Prisms do not have an apex, this is true for a pyramid.13. T14. F: All pyramids have triangular faces, not rectangular.15. F: A sphere is not comprised of polygonal faces.16. T17. T18. F—1 gallon � 3.784 liters.

Multiple Choice.19. d 20. d 21. c 22. e 23. e24. a 25. d 26. d 27. a 28. b29. b 30. a 31. e

Fill in the Blanks.32. Diameter33. ray34. complementary35. hypotenuse36. pentagon37. rectangular regions38. (a) Solid: Tetrahedron Shape of faces: Triangle Number of faces: 4

(b) Solid: Cube Shape of faces: Square Number of faces: 6(c) Solid: Octahedron Shape of faces: Triangle Number of faces: 8(d) Solid: Dodecahedron Shape of faces: Pentagon Number of faces: 12(e) Solid: Icosahedron Shape of faces: Triangle Number of faces: 20

39. Base is: Regular HexagonLateral Faces Scalene Triangles

Writing.40. Congruent sides and parallel sides should be mentioned.41. Squares and rectangles have all right angles

Squares and rhombi have all four congruent sides Parallelograms have opposite sides parallel Trapezoids have exactly one pair of parallel sides and if the two nonparallel sides are congruent, it is an isosceles trapezoid Kites have two pairs of congruent sides.

DEAB

trapezoids

non-parallelograms

kitesisosceles trapezoids

parallelograms

rhombusrectangles squares

42. Congruent angles, congruent sides, number of angles, sides, and lines of symmetry should bementioned.

43. Regular triangles because 60� evenly divides into 360�; squares because 90� evenly divides into360�; regular hexagons because 120� evenly divides into 360�.

Exercises/Problems44. (a) 90� � 30� � 60�, 60� � 23� � 83�, 180� � 83� � 97�, �DAE � 97�

(b) �GAD and �DAF, �GAC and �CAF, �FAB and �GAB,(c) and are the only complimentary angles in this figure.(d) There are many, here are five pairs: �GAC and �CAB, �CAB and �BAF, �BAF and

�FAE, �FAE and �EAD, �EAD and �DAG45. (GET CORRECT MINUTE SYMBOL)

(a) 43�42�(b) 113�40�(c) 56�24�(d) 29�7�

46. , , and 47. (a) 3.05

(b) 3.65(c) 1.1(d) 5.6

48. �X � 124�; obtuse.49. 135�50. 53�51. (a)21 (b) 4 (c) 5 (d) 6 (e) 1052. (a) �DEF, �QRT

(b) �ABC, �JKL, �MNP, �GHI(c) �DEF, �MNP(d) �GIH(e) �JKL, �QRT

53. 40�54. An isosceles trapezoid.

55. Any “stretched” out hexagon will work.

56. Divide the pentagon into three triangles. The sum of the degree measures of the three triangles is the same as the sum of the degree measures of the vertex angles of the regular pentagon. Since each vertex angle in the regular

pentagon is the same, 5v � 180� � 3; v � . Since 3 � n � 2 for

n � 5; in general you can see that for a regular n-gon, the measure

of the vertex angle is v � .180° * (n - 2)

n

180° * 35

BE!

BD!

BC!

BA!

EABA


57. X � 48�, Y � 32�58. 30�59. F � 6, V � 6, E � 10, so F � V � E � 2 6 � 6 � 10 � 2 12 � 1260. (a) �ABC, �DEF

(b) ABFD, BFEC, ACED(c) �DEF, ACED, ABFD(d) , , , , , , , ,(e) right triangular prism

61. A right regular pentagonal prism.62. 432 minutes63. 4498.848 m/min64. (a) 13.95 miles per hour

(b) .23 miles per minute(c) 22.45 kilometers per hour

Applications.65. 13.26°66. 142�67. 43�, 106�68.

69.(a) There are many, here are some examples

(b) There is only one such square

A

B

A

B

DFDEADABACCBBFEFCE

QQ


(c) This is not possible, the only way to connect lines, on the grid, forming right angles at Aand B, is to form a square.

(d) There is only one possible isosceles trapezoid that fits on this triangular grid

70. (a) There are two nontrivial rotational symmetries about point A. 120� and 240�(b) There are no lines of symmetry—there is no way to sketch a line that you can fold the

figure in half around and match the opposite sides.

71. (a) There are five nontrivial rotational symmetries about point A. 60�, 120�, 180�, 240�and 300�

(b) There are six lines of symmetry as shown here:

72. (a) �CHD & �GHD, �CHD & �AHC, �AHC & �AHG, �AHG & �DHG(b) A hexagon(c) 127�(d) 41�(e) �AHC � 102�, �GHD � 102� and �BCH � 152�(f) Example: ABDEGA(g) Example: AHG and GHE(h) �DFE and �DEF, �DHG and �CHA, �CHD and �AHG,

73. 43� 25�74. 68� 46� 4575. It is short 6 feet.

A

A

B

A

B


76. Three rectangles and 2 triangles

77. N-gon pyramids have 2n edges; n � 14, n � 1 vertices, V � 15, n � 1 faces F � 15. Check withEuler’s formula: 15 � 15 � F � V � E � 2 � 30.(a) n � 14(b) F � 15(c) V � 15

78. N-gon prisms have 2n vertices; n � 32, n � 2 faces, F � 34, 3n edges E � 96. Check with Euler’sformula: 34 � 64 � F � V � 96 � 2 � 98.(a) n � 32(b) F � 34(c) E � 96

79. Two circles and one rectangle

80. 75.69 1/sec81. $14.6382. 2178 light years83. 13.86 km/l


GEOMETRY Chapter Three—Answers

True–False.1. T2. F: There are many counterexamples3. F: There are many counterexamples4. F—The perimeter � 8 inches.5. T6. T7. T8. T

9. F—The area of the square � s2 the area of the triangle �

10. F: The triangle must be a right triangle.11. T12. F—Surface area is expressed in square units.13. T

14. F—V � .

15. F—V � �r3.

16. T

17. F: It is of the volume of a right circular cylinder with the same diameter.

Multiple Choice.18. a 19. b 20. d 21. b 22. c23. e 24. d 25. c 26. d 27. b28. a 29. b 30. a 31. c 32. a

Fill in the Blanks.33. pi34. square

35.

36.37. base radius; slant height38. 3:r39. 2A � (n � a) � h

40. �r3

Writing.41. A picture may be used to show that the areas are in a ratio of 2:142. The circle will be inside the square. The area of the square is 1 unit � 1 unit � 1 unit2 and the

area of the circle is �r2. Since C � 1 unit, r � 1/2 unit and �(1/2)2 � �(.25) � .79.The area of the

circle � .79 unit2 which is smaller than the area of the square.

21 unit

1 unit

43

2n2 + m2

12

(n * a) * h

23

43

13

Bh

14

s2 23

43. If you drop a perpendicular bisector from one vertex of the equilateral triangle to the oppositeside you have a special 30°–60° right triangle. Use the Pythagorean Theorem to see:

44. Because the triangle is 45°–45° the leg lengths are both s. Use the Pythagorean Theorem tosee:

Exercises/Problems.

45. The area of the circle is � times greater than the area of the square.The perimeter of the circle

is � times greater than the perimeter of the square.

46.

47. a. 20 cmb. 17 feetc. 16.8 metersd. 12.35 cm

48. a. 4.5 cm2

b. 10 cm2

c. 8.61 cm2

d. 112 cm2

49. 7.7 cm50. 37.70 ft51. 60.24 sq. cm52. 294 sq. in.53. Area � 26.65 sq. cm; perimeter � 24.92 cm

54. cm and cm

55. 657.02 sq. in.56. 1536 sq. mm57. 301.59 cu. in.

7226

7222

.b .a

1.5 sq units

1 sq unit

2 sq units

1 sq unit

2 sq units

1 sq unit

1 sq unit 2 sq units

2 sq units

2 sq units

2 sq units

4 sq units

4 sq units

14.5 u2 u 41 2

12

90° 45°

45°

d

s

ss2 + s2 = d2; 2s2 = d2; d = 22s,

60°

60° 60° 60°

30°

2

s s

s

h

s2 - a s

2b2

= h2; 3s2

4= h2; h =

23s

2


Applications.58. 57.6 acres59. 3n � 2 units60. 233,734,493.4 miles61. 34 tiles62. 31%

63. Perimeter � (.75� � 1.5) units, area � units2

64. 116.7 feet2

65. 3.8 cm2

66. 12� feet2

67. R � � 16.9768. (a) 50.27 sq. in. (b) 30%69. 100.6 ft.70. 125.59 miles71. 11.4 feet72. d � x73. 131.76 sq. ft.; 48.76 sq. ft.74. 26 times larger75. 28.35 sq. ft.76. a. Surface area � area of base � area of six triangles � � cm2

b. Volume � � 8 cm3

77. 198� cm2

78. 186,349 times larger79. 109 board feet80. 180� cm3

13

* 5423

182915423

23

2288

234

+3p32


GEOMETRY Chapter Four—Answers

True–False.1. T2. F: This is the converse of the correct statement p r.3. F—”if p then q” is the same as “p only if q.”4. F—A statement and it converse may or may not have the same truth value.5. T6. F: There are six pairs of congruent parts.7. T8. F: SSA is not a valid congruence condition.9. F: By Identity

10. F— must be perpendicular to to reach the conclusion.11. T12. T13. T

Multiple Choice.14. e 15. b 16. d 17. e 18. a19. d 20. d 21. f 22. g 23. d24. e 25. c 26. a

Fill in the Blanks.27. Possible conclusions: not r, not q or not p.28. if and only if29. corresponding parts of congruent triangles are congruent30. �M31. The reflexive property.32. equiangular33. �

Writing.34. Answers will vary. One possible solution is:

If you get enough sleep then you can study effectively (p q)If you study effectively then you will do well on your exam (q r)Using the Law of Syllogism, we have p r, which is the statement:If you get enough sleep then you will do well on your examUsing the Law of Detachment, we have, if p then r; p, therefore r. The contrapositive of this isIf not r then not p, which is the statement:If you did not do well on your exam then you did not study effectively

35. Answers will vary. An example might be:If the sun shines, then the lawn dries out.If the lawn dries, then the sprinklers start.If the sprinklers start, then water gets on the road.If water gets on the road, then the road gets wet.

36. Answers will vary. One possibility is:HL refers to Hypotenuse-Leg, which means the two triangles under consideration are bothright triangles. Therefore we can use the Pythagorean Theorem to show the other twocorresponding legs are also congruent and apply either SAS or SSS.

QQ

Q

BPAP

ABPQ

Q

37. Answers will vary but should indicate two triangles with three congruent angle pairs but different length sides. For example: �A � �X, �B � �Y and �C � �Z but is not congruent to .

38. Answers will vary. Properties which might be included:Two sides of the triangle are congruent.Two angles of the triangle are congruent.Median to the base bisects the base.Median to the base is perpendicular to the base.Median bisects the vertex angle.Median separates triangle into two congruent triangles.

Exercises/Problems.39. No conclusion is possible40. Possible conclusions include:

SOFT has four congruent sides.SOFT has four congruent angles.SOFT is a rhombus.All squares are rhombuses.

41. �ABC is a right triangle.42. No deduction possible43. True; if �A � �C, then ABCD is a parallelogram. The converse is false.44. Possible conclusions include:

not tnot s

45. The statement is true.Converse: If a2 � b2 � c2 then this is a right triangle with sides a, b and c and c the longest side.Biconditional: A triangle with sides a, b and c, with c the longest side, is a right triangle if andonly if a2 � b2 � c2.

46. The statement is true.Converse: If two angles are complimentary then the sum of two angles is 90�.Biconditional: The sum of two angles is 90� if and only if the angles are complimentary.

47. �P � �M, �A � �N, �T � �R, � , � , �48. �AGF � �EGF by ASA49. Pair One: �ADB � �CEB by SAS

Pair Two: �DAB � �ECB by CP� by CP

�AEB � �CDB since �DEB is an isosceles triangle� by the reflexive property, therefore � by adding two congruent line segments

�AEB � �CDB by SAS50. �AEC � 67.5�, �CAB � 30�51. (a)

(b)(c) �ADE

52. 67�53. (a) 54�

ACAB

EADCEDDE

CBAB

MPPTNPATMNPA

X

Z

YA

C

B

XYAB


(b) 108�(c) 18�

54. 38�55. Compass marks should look approximately like those in the following diagram.

56. Compass marks should look approximately like those in the following diagram.

57. One way:

58. Compass marks should look approximately like those in the following diagram.

59. Use Construction 4.60. Use Construction 1 twice (or three times).61. Construct a perpendicular to a line forming a 90� angle. Bisect it to get the 45� angle. Then

construct an equilateral triangle (any size) to get a 60� angle. Then 45� � 60� � 105�

62. Construct a perpendicular at an endpoint of the hypotenuse. Bisect the hypotenuse to get c.See below.

63. Use Construction 6 to find the midpoint of .BC

C

1/2 C

12

Construct perpendicular Bisect to form 45° angle Bisect to form 22.5° angle

Construct perpendicular Bisect to form 45° angle Add 60° angle to the 45° angle

C

B

A

ED

B

FC

B


Proofs.64. (Outline) , �ABD and �CBD are right triangles, and B the midpoint of . Thus,

� . DB � DB. So �ABD � �CBD by LL.65. (Outline) �ABC and �DCB are right triangles and � , so �ABC � �DCB by LL.Then

�BAC � �CDB by C.P.66. (Outline) �CAB � �CBA by Thm 4.5, �FAB � �FBA by Thm 4.5. �FBA � �FBD �

�DBA, �FAB � �FAE � �EAB therefore �FBD � �EAE. � given. �BCA ��ACB identity. �EBC � �DAC by ASA.

67. �B � �C since they are opposite congruent sides. Also, their sum is 120�. Hence �A � �B ��C � 60�, so �ABC is equilateral.

68. (Outline) Since �RVT is isosceles with �V the vertex angle, � . Then, in �RVT, �R ��T since they are opposite congruent sides. is the median to � . Then �RVS� �TVS by SAS. By C.P., �RVS � �TVS. Thus bisects �RVT.

69. � since they are opposite sides in a parallelogram. � and � becauseand bisect each other. �AGB and �EGD are congruent because they are vertical

angles. So �AGB � �DGE by SAS. � by C.P. Finally, since AB � BC � AC and FE �ED � FD, � .

Applications.70. This traverse closes if and only if the sum of the measures of the angles in it is 180�.

30� 46� 3 � 70� 14� 40 � 78� 59� 17 � 180�. So it closes.71. The four triangles are equilateral and congruent, which can be proved by SAS. Since the cuts

are made at the midpoints of the sides, the new sides are one-half the original length.72. The point of intersection of the perpendicular bisectors of and will be the point which

is equidistant from each of the three points.ACAB

FEBCEDAB

BEADGEBGGDAGFDAC

VSSTRT Q RSVS

VTRV

BCAC

BCBCBCAB

ACAC�DB

CM B

A


GEOMETRY Chapter Five—Answers

True–False.1. F: There are many counterexamples including three parallel lines and three lines with no

parallel relationships.2. T3. F: Corresponding angles are congruent if the lines are parallel.4. F: The two interior angles must be nonadjacent.5. T6. T7. T8. F—They are perpendicular.9. T

10. F: Only if it is an isosceles trapezoid11. F: It depends on the position of points A, B, C and D.

12. T13. T14. T

Multiple Choice.15. d 16. e 17. d 18. b 19. c20. d 21. a 22. c 23. d 24. d25. b 26. a 27. e

Fill in the Blanks.28. (a) Alternate interior angles

(b) Alternate exterior angles(c) Interior angles on the same side of the traversal(d) Exterior angles on the same side of the traversal(e) Corresponding angles

29. parallel; skew30. parallelogram31. OTRI is a parallelogram32. congruent; perpendicular

Writing.33. Answers will vary. Assume p and not q to be true. Deduce a contradiction.34. Answers will vary. (See last page of section 5.4 for a summary table.)

Exercises/Problems.35. (a) �2 and �15; �5 and �14

(b) �8 and �12; �3 and �13(c) �6 and �14; �1 and �15(d) �4 and �13; �7 and �12(e) There are actually four labeled pairs: �7 and �14; �6 and �12; �4 and �15; �1 and

�13

DC

BA

36. (a) 39°(b) 103°(c) 38°

37. 35°38. (a) False

(b) True: The exterior angle equals the sum of the non adjacent interior angles.(c) False(d) True: Alternate interior angles are congruent.

39. (a) 95°(b) 25°(c) 48°

40. �2 and �15; �5 and �14 by alternate interior angles �7 and �14; �6 and �12; �4 and �15;�1 and �13 by corresponding angles.

41. (a) 180°(b) �2 and �4(c) 50°

42. (a) 31°(b) 59°

43. (a) True, �4 � �6 and �6, �9 and �13 are the three angles of a triangle.(b) False, �5 is the supplement of �8 � �10(c) True, �1 � �14 and �4 are both supplements of �5(d) True, sum of all of the angles of two triangles.

44. 1080 sq. cm45. (a) True for both parallelograms and rhombi. Diagonals bisect each other in all

parallelograms(b) F(c) True for both parallelograms and rhombi. Opposite sides are congruent in all

parallelograms(d) True for rhombi, but not for non-rhombi parallelograms. All sides are congruent in a

rhombus.46. (a) True for rhombi, but not for non-rhombi parallelograms. Diagonals are perpendicular in

a rhombus.(b) True for both parallelograms and rhombi; vertical angles.(c) F (supplementary, not congruent)(d) True for rhombi, but not for non-rhombi parallelograms. Diagonals bisect the opposite

angles in a rhombus.47. �1 � 60°, �2 � 30°, �3 � 60°, �4 � 90°48. �QRP � �SPR, �SPQ � �QRS, �STR � �QTP, �STP � �QTR49. (a) True for both rectangles and squares; opposite sides are congruent.

(b) True for all parallelograms, so true for both rectangles and squares; diagonals bisect eachother.

(c) True for squares, diagonals are the same length, so are of the diagonal.(d) F

50. (a) True for square (which is a rhombus), but not for non-square rectangle. Diagonals areperpendicular in a rhombus.

(b) True; vertical angles(c) True for square (which is a rhombus), but not for non-square rectangle. Diagonals are

perpendicular in a rhombus so this angle is 90°.(d) True for square; diagonals bisect vertex angles.

12


51. 30.30 sq. in.52. (a) 67°

(b) 23°(c) 175 sq. meters

53. Use Construction 8.54. Use the result that the diagonals of a rhombus are perpendicular bisectors of each other.55. (One possible outline) Construct two perpendicular lines, copy length a on both lines, swing an

arc of length a from each end point to form the fourth vertex of the square.56. Use the idea that any segment from the base of the triangle to a line parallel to the base and

through vertex A will be an altitude of �ABC as well as �ABM1 , �ABM2 and �ACM3. Withthe same altitude it is sufficient to use Construction 8 to form points M1 and M2.

Proofs57. (Outline) �BDC � 60° and implies that �CBD is a 30–60 right triangle. The

hypotenuse is and the side opposite the 30° angle is , so CD � 2BD.58. (Outline) �BAC � �ACD (alternate interior angles) and � . So �ABC � �CDA by

AAS. Then �BCA � �CAD by C.P. and then , since alternate interior angles arecongruent.

59. (Outline) In the figure W, X, Y and Z are the midpoints. �B � �D and �A � �C becauseopposite angles in a parallelogram are congruent. � � � and � �

� because � and � in the original parallelogram. Then �WBX ��ZDY and �WAX � �XCY by SAS. Thus, by C.P., � and � so WXYZ is aparallelogram.

60. (Outline) TQ // SR (Given). � (Identity), � � � (HL), �STU � �SPU(C.P.), �SPU � �RQS (Corresponding angles), �STU � �RQS (transitive), therefore TQRSis an isosceles trapezoid since its base angles are congruent.

61. Since �BFE � �BEF, � . Then since the trapezoids are isosceles, � � �. ABFG and BCDE are parallelograms because they each have a pair of opposite sides

which are both parallel and congruent.62. �DAC � �BAC because the diagonals of a rhombus bisect its angles. � since all the

sides in a rhombus are congruent and � , so �ADC � �ABC by SAS.Then by C.P.� so ABCD is a kite since it has two pairs of consecutive sides congruent.CDCB

ACACABAD

CDBFBEAGBEBF

TUSPUSUSUS

B X C

A Z D

YW

WZXYZYWXCDABADBCZAWA

YCXCYDZDXBWB

BC‘ADACAC

BDCDBD�CA

M 2M1 C

B

A


Applications.63. 4882 km.64. The sun’s rays are roughly parallel so that the angles formed at P and Q are congruent (They

are corresponding). The trees are assumed to be at right angles with the ground and theshadows are the same length. So the triangles formed are congruent by AAS and thus, by C.P.,the trees must be the same height.

65. The storage facility should be built at the point of intersection of the two angles bisectors asshown.

66. The crossed legs form diagonals of a quadrilateral and since they intersect at their midpoints,the quadrilaterals formed must be parallelograms. Thus the seat and the ground, as “oppositesides,” will be parallel.

67. Yes. See the sketch. The angle of incidence equals the angle of reflection so that a series ofparallelograms is made by the path of the ball.

68. Actually the quadrilateral does not need to be a parallelogram for this to be true. The sum ofthe angles in any quadrilateral is 360° and so they will fit perfectly in one rotation about a point.

69. The base of the tree being at the center of an equilateral triangle means that it is equidistantfrom the vertices.The tree is held perpendicular to the ground so the three triangles formed bythe tree, the wires and the distance from the tree to the vertices of the equilateral triangle arecongruent by LL. The length of each wire should be approximately 3.8 feet.

70. 1374.7 cu. ft.

60°

60° 60°1 3

4

52

Q

P


GEOMETRY Chapter Six—Answers

True–False.1. F—It is a proportion.2. T3. F—The mean proportional is .4. T5. F: There are many counterexamples.6. T.7. T8. T9. F—Every triangle has three midsegments.

10. F—tan X � .

11. T12. F—(tan A)(cos A) � sin A.13. T14. F: The Law of Cosines works on any triangle, not just right triangles.

Multiple Choice.15. d 16. a 17. c 18. b 19. d20. a 21. a 22. e 23. d 24. e25. f 26. e 27. a 28. C

Fill in the Blanks.29. Terms, extremes, means.

30. if and only if ad � bc.

31. The geometric mean (mean proportional).32. ratio; proportion33. similar34. congruent35. complementary

Writing.36. Answers will vary.

AA: Two angles of one triangle are congruent to two angles of a second triangle.SAS: Two sides of one triangle are proportional to two sides of a second triangle and theincluded angles are congruent.SSS: Three sides of one triangle are proportional to three sides of a second triangle.

37. Answers will vary. Possibilities include:�A and �B are complementary.sin A � BC/AB; sin B � AC/AB; cos A � AC/AB; cos B � BC/AB.tan A � BC/AC; tan B � AC/BC.sin A � cos B; cos A � sin B; tan A � (sin A)/(cos A).sin2 A � cos2 A � 1.

Exercises/Problems.38. Law of Sines example should be a triangle with the measure of two sides and an angle opposite

one of them or the measure of at least one side and any two angles given.Law of Cosines example should be a triangle with the measure of all three sides or themeasure of two sides and their included angle given.

a

b=

c

d

RP

RX

1mn

39. 840.41. X � 3.5 cm, Y � 1.725 cm. �R � 70�.42. CD � 24.43. BD � 9.44. (a) �FOL � �DSL by AA

(b) �TRI � �ODP by SSS Similarity Theorem45. 36/16946. (a)

(b) 2.625(c) 3:8

47. (a) 14 cm(b) 20 cm(c) 57�(d) 42�

48. 16.2 sq cm49.

50.

51.

52. �B � 41�, b � 8.6 and c � 13.1.53. �B � 63.2, �C � 66.8� and b � 11.7.54. �A � 51.3, �B � 59.2 and �C � 69.5�.55. �A � 44.6, �B � 50.4 and c � 14.2.

Proofs.56. Answers will vary.

(Outline)

57. Answers will vary.(Outline) Since the rectangles are similar, A�B� � x(AB) and B�C� � x(BC). The cylindersformed will have base radii of AB and A�B� and heights of BC and B�C�.Thus the volumes willbe V1 � �(AB)2(BC) and V2 � �(A�B�)2(B�C�) � �[x2(AB)2][x(BC)] � �[x2(AB)2][x(BC)] �x3 �(AB)2(CD) � V1x

3. Thus, since x is the ratio of the corresponding sides, the ratio of thevolumes is the cube of x.

b

a + b=

d

c + d

a

b=

c

dQ ad = bc Q ad + bd = bc + bd Q d(a + b) = b(c + d) Q

5.6 feet

4.2 feet53°7 feet

A

T

C37°

55.2

34.8°

33 feetC

T

S

7 feet4 feet

51.3°

38.7°

41 feet4 feet

5 feetA

B

C

623

1023


58. (Outline) Since L and M are parallel, �ACE � �ABD (corresponding angles). Since �ACEand �ABD are right triangles, �ACE � �ABD by AA similarity.x � 6 cm, L � 2 cm, M � 5 cm,

59. Answers will vary.(Outline) �ABC � �BDC so BC:AC � DC:BC.Then (BC)2 � (AC)(DC) BC � . So BC is the mean proportional of AC andDC.

60. Answers will vary.(Outline) �APQ � �AQR � �BAR so that AP:PQ:AQ � AQ:QR:AR � AB:AR:BR andAP:PQ � AB:AR.Thus, AP/PQ � AB/AR. But AP/AR is the sine of �AQP and AB/AR is thecosine of �BAR. So sin(�AQP) � cos(�BAR).

61. By the Law of Cosines; cos B � , cos2 B �

By right triangle trigonometry, sin B � . Since sin2 B � cos2 B � 1 and sin B �

we have that h � a and c � h � ac

as desired.

Applications.62. 87.5 km; 54.4 mi.63. 4.2664. 96�65. 12�66. 15�67. 25�68. 711.5�69. 125.7�70. 369 sq. ft.71. Approximately 59.8 feet2.72. Approximately 17.2 feet3

73. Approximately 48.7 feet.

A1 - aa2 + c2 - b2

2acb21

212A1 - aa2 + c2 - b2

2acb2

21 - cos2Bha

aa2 + c2 - b2

2acb2a2 + c2 - b2

2ac

2(AC)(DC)Q

2525


GEOMETRY Chapter Seven—Answers

True–False.1. T

2. F–Area � sq. in.

3. T4. T5. F–�MPX� 0.5( � ).6. If two chords intersect, the measure of any one of the vertical angles formed is half the sum of

the two arcs intercepted by the two vertical angles.7. T8. F: The measure of an angle formed when a chord intersects a tangent line (at the point of

tangency) is half the measure of the arc intercepted by the chord and the tangent line.9. T

10. F—It is a right triangle.11. T12. F—It is the centroid.13. T

Multiple Choice.14. d 15. e 16. d 17. b 18. c19. a 20. a 21. a 22. e 23. d24. c 25. b 26. c 27. b

Fill in the Blanks.28. perpendicular bisector of 29. The points A and B of a circle together with the portion of the circle between the two point.30. Semicircle31. Minor arc32. Major arc33. the chords are not parallel34. perpendicular35. perpendicular bisectors of the sides; circumcenter


When secants intersects inside a circle, (AE)(BE) � (CE)(DE) because �BEC � �DEA byAA. Then the sides are proportional.

C

A B

D

E

AB

NY¬MX¬

50p3


When secants intersect outside a circle, (BE)(AE) � (CE)(DE) because �ACE � �DBE byAA. Then the sides are proportional.

37. The radius of a circle is a segment from the center of the circle to any point on the circle.A chord is any line segment whose endpoints are on a circle.A diameter is a chord that goes through the center of a circle.A secant line is a line that intersects a circle at two points.A tangent line is a line that intersects a circle at exactly one point.

38. Answers will vary.Since the set of points, A, B, C and P are an orthocentric set, A is the orthocenter of �BCP; Bis the orthocenter of �ACP; C is the orthocenter of �ABP and P is the orthocenter of �ABC.

Exercises/Problems.39. �DEB and �DAB, both measure 50�.40. 45�.41. �EDA and �ABE are both inscribed and �ACE is central.42. Arcs ED, DB, BA and AE are all minor arcs.

Arcs EDA, EAD, DBE, DEB, BAD, BDA, ADB and ABD are all major arcs.Arcs EDB, EAB, AED and ABD are all semicircles

43. (a) 80�(b) 40�

(c)

44. (a) 10(b) 30�(c)

45. 120�46. 4 cm or 10 cm47. (a) 10

(b) 7.5(c) 32/3

50p - 5013 sq. in.

329

p cm

Tangent

Chord

SecantDiameter

Radius

B

C DE

A


48. (a) 31�(b) 101.5�(c) 47.5�(d) 90�

49. Arc AD by the Inscribed Angle Theorem.

50. �BEC � (arc AD + arc BC)

51. (AE)(CE) � (BE)(DE)

52. �AED � (arc AD � arc BC)

53. (AE)(BE) � (DE)(CE)

54. �ADC � (arc AC � arc BC)

55. (DC)(DC) � (DB)(BA)

56. �ABC � (arc ADC � arc AEC)

57. AB � CB58. Answers may vary.

Construct the perpendicular bisectors of two sides to find the center of the circle. The radius isthe distance from this point to any of the three vertices.

59. Answers may vary.Construct the angle bisectors of two of the angles of the triangle.This is the center of the circle.Construct the perpendicular from this point to any of the three sides of the triangle. Thisdistance is the radius of the inscribed circle.

60. Answers will vary but should consist of creating the perpendicular bisectors of two arbitrarynonparallel chords on the circle. The center will be the intersection of the two bisectors.

61. Answers will vary but should consist of creating the perpendicular bisectors of two sides of thepentagon, using the intersection of the two bisectors as the center of the circle and drawing acircle using this center and on of the vertices of the pentagon to mark the radius of the circle.

62. Answers will vary.To construct the tangent at P, construct the perpendicular to at P.To construct the tangent from Q, construct the midpoint, M, of . Construct the circle withcenter M and radius OM. Label the point of intersection of this circle and the original circle as A.Construct .


�DAC � �DBC (Angles inscribed in same arc) and�AED � �BEC (Vertical angles), so�DAE � �CEB (ASA). Then � by C.P.

64. Answers will vary.See sketch. Since bisects , � . � and � (radii of same circle).Then �AOD � �BOD (SSS). Thus, by C.P. �AOD � �BOD. But � �AOC and

� �BOC so � . Then the line bisects the arc .AB¬ÍOC

!CB¬AC¬CB¬

AC¬OBAOODODDBADABOC

BCAD

QA

OQOP

12

12

12

12


65. Answers will vary.Since AD � DC and O is the center of the circle, � . � (radius and tangent areperpendicular). So �OCP is a right triangle and is the altitude to the hypotenuse.Then CDis the geometric mean of OD and PD.

66. Answers will vary.Draw the perpendicular from O� to , and call it O�B. Then since and are bothperpendicular to , PP�O�B is a rectangle and is congruent to �OO�B is a righttriangle with hypotenuse r � R. OB � R � r. By the Pythagorean Theorem, O�B �

. But PP� � O�B�, so PP� � .

67. Answers will vary. See sketch.�PTR is a right angle since is a semicircle. Then �PTR is a right triangle with thealtitude to the hypotenuse. Thus TQ is the mean proportional (geometric mean) of PQ and QR, and TQ � � � .

T

PQ M R1

a

a

2a2(1)(a)2(PQ)(QR)

TQPTR¬

P

Br

r

O

R-r

P�

O�R+r

22rR2(r + R)2 - (R - r)2 = 24rR = 22rR

O¿B¿PP¿PP¿O¿POPOP

O D

C

P

CDPCOCACOP

OD C

A

B


Applications.68. 455 cu ft69. Since C is the center and A, B, D and E are all on the circle, CA, CB, CD and CE are all radii

and therefore congruent. �BCD is 60� (vertical angles). Thus �BCD is an isosceles triangleand �CDB � �CBD � 60�.

70. Using special triangles;

71. 14%; 21.55 sq. in; 50.4�72. 5807 miles73. 13 cm74. 731�75. r � 8.5 miles; area � 221.58 sq mi76. 1.25 miles77. 6; 102.9 sq in.78. Compass marks should show finding the centroid of the triangle.

a a16p *13b - a1

2* 423 * 2b b cm2 L 9.83 cm2


GEOMETRY Chapter Eight—Answers

True–False.1. T2. F—AC � � .3. T4. T5. F; only if m 06. T7. T8. F: The center is not on the circle.9. F—It may also have no solution or infinitely many solutions.

10. T11. F—The fourth vertex is at (a, b).

Multiple Choice.12. b 13. e 14. a 15. a 16. d17. c 18. e 19. d 20. b 21. e22. d 23. a 24. d

Fill in the Blanks.25. x-axis26. y-axis27. C is between a and B28. The product of the slopes � �1.29. are parallel30. Are the same31. Intersect (once)32. a rectangle


Positive slope: slants up from left to rightNegative slope: slants down from left to rightZero slope: horizontalUndefined slope: vertical

34. Parallel lines have the same slope.In the equations y1 � m1x � b1 and y2 � m2x � b2, m1 � m2.Examples: y � 2x � 3, y � 2x � 4If m � 0, both parallel lines will be horizontal and of the form y � b.Examples. y � 3 and y � 4.Perpendicular lines have slopes whose product is �1.In the equations y1 � m1x � b1 and y2 � m2x � b2, m1 � m2. � �1

Examples: y � �3x � 3, y � x � 4

If one line has slope 0, it will parallel to the x-axis and of the form y � b. Any line of the formx � c will be perpendicular to lines of the form y � b (undefined slope).

35. Answers will vary.One solution: two intersecting linesNo solution: parallel linesInfinitely many solutions: same line

13

Z

1317

Exercises/Problems.36.37. (a)

(b)(c) (1/2, 1)

38. b,d39. Isosceles; PQ � QR � and PR � ; area � 3040. No. Slope of � �2/3 and slope of � 4/3.41. a � 17/242.

(a) y � .

(b) m � ; y intercept � (0, 2)

(c) 2(�1) � �3(2) � 4

43. .

44.

45. (x � 0)2 � (y� 1.5)2 � (6.5)2

6

4

2

5

(6, -4)

(-6, 1)

(0, -1.5)-2

-4

-5

-6

-8

a 211

, - 2511b

y =23

x -73

- 32

- 32

x + 2

QRPQ415165

1113

A

Y

X

B

(3, 213 + 1)


46. (a) y � x � 1

(b) x � �4(c) y � �2

47. x � 2 y � 1348. No solution49. Midpoint of � midpoint of � (4, 2). Slope of � �5/3, and slope of � 3/5

50.

51.

52.

53. (0, 1)54. (�1, 0)


M(2m, 2n); PM � and PQ � so PM � (1/2)(PQ)56. Answers will vary.

Slope of AB � n/m and slope of AC � �m/n57. Answers will vary.

The center of the circle is (a, a) and its radius is a so the points of tangency are (a, 0) and (0, a).58 Answers will vary.

One possibility is shown below. Length of hypotenuse � and the midpoint of the

hypotenuse is M(a/2, b/2), so AM � � one-half the length of the hypotenuse.

59. Answers will vary.One possibility is shown. AC � BD �

A (-a, 0)

D (-b, c)

B (a, 0)

C (b, c)

2a2 + 2ab + b2 + c2.

A (0, 0) B (a, 0)

C (0, b)

122a2 + b2

2a2 + b2

22m2 + n22m2 + n2

a -12

, -132b

a132

, 12b

a122

, -122b

BDACBCAC

23


60. Answers may vary.The midpoint of � N � (1, 4). Then the line through C and N is y � �6x � 10

The midpoint of � T � (2, 8). Then the line through A and T is y � x � 5.

The point of intersection, M, of these two lines is (2/3, 6) Then AT � and AM � .

Applications.61. 19,200 acres62. 5/8; 5 in 863. 0.51; 3.63�64. 7.2; 57.6; 8465. (a) $2062.50; $2125

(b) C � 1.25w � 2000(c) 1.25(d) $11,295

66. (a) 36.6 57(b) S � 3.4d � 23(c) 3.4(d) 74

67. Circle 1: (x � 5.5)2 + (y � 5.5)2 � (1)2

Circle 2: (x � 5.5)2 + (y � 5.5)2 � (2)2

Circle 3: (x � 5.5)2 + (y � 5.5)2 � (3)2

Circle 4: (x � 5.5)2 + (y � 5.5)2 � (4)2

Circle 5: (x � 5.5)2 + (y � 5.5)2 � (4.5)2

68. Line B: x � 5.5; Line D; y � 5.569. �(4.52 � 42) � 4.25� feet2

70. Answers will vary. One way: D((3, 4), (5.5, 5.5)) � .The point (3, 4) is in the white area in the 3rd ring just under Line D

71. Using 45�–45� triangles. Line A: y � �x � 11; Line C; y � x

72.

21.52 + 2.52 L 2.9

23152152

32

BC

AB


GEOMETRY Chapter Nine—Answers

True–False.1. F—It is called a transformation.2. T3. Reflections and Glide Reflections do not.4. It is an isometry, not just a rotation.5. T6. T7. F—The areas are equal.8. T9. T

10. T11. F—The ratio of the areas is 25.12. T13. The reflection component of a glide reflection does not have a fixed point.14. F: Size transformations have the identity transformation: Center 0, scale factor 1.15. F: The polygons must be regular for this to occur.16. F: if a component of the similitude is a size transformation with scale factor 1.

Multiple Choice.17. e 18. c 19. a 20. d 21. c22. e 23. c 24. a 25. b 26. b27. a 28. e 29. e 30. d

Fill in the Blanks.31. An isometry32. A glide reflection33. line; vector (Order may be reversed.)34. congruent35. M36. identity size37. similitude


Translation: requires a vector or directed line segment.Rotation: requires a point (center) and a directed angle.Reflection: requires a lineGlide Reflection: requires a vector and a line

39. Rotations: For all regular n-gons there are n rotations about the center of the polygon with

directed angle the 1, . . . , n multiples of . For example, for a triangle, the directed angles

for the rotations would be 120�, 240� and 360�. For a square: 90�, 180�, 270� and 360�.

360°n

Z

Reflections of even sided regular n-gons: If a regular n-gon has an even number of sides there

are reflections through the lines that bisect the opposite sides of the polygon and

reflections through the lines from opposite vertices for a total of n reflections. The square andregular hexagon are given as examples.

Reflections of odd sided regular n-gons: If a regular n-gon has an odd number of sides thereare n reflections through the lines that bisect a sides of the polygon and go through theopposite vertex. The equilateral triangle and regular pentagon are given as examples.

40. Answers will vary.Preserves linearity, angle measure, perpendicularity, parallelism . . . these shared with isometries.Preserves ratios of distances and orientation.

Exercises/Problems.41. (a) A�(6, 3) and B�(1, �5)

(b) A�(1, 2) and B�(�4, �6)42. Construct so that ABB�A� is a parallelogram with the length of and the same

length as the vector. Follow the same procedure to get the image of C�.43.

44. Answers may vary. Draw and use the protractor to find A� so that �A�OA � 80� (acounterclockwise turn) and OA � OA�. Do the same for B� and C�.

45. Answers may vary.Construct so that AA�B�B is a parallelogram with the side the same length as .This accomplishes the translation.Then construct the perpendicular through line 1 from A� andfrom B� and mark A and B on these perpendiculars so the distance from line 1 to A� and fromline 1 to A are the same. That is, line 1 forms the perpendicular bisector of and of .B¿B–A¿A–

PQAA¿A¿B¿

OA

C

C�

B�

A�O

A

B

AA¿BB¿A¿B¿

Lines through midpoint of side and opposite vertex

Opposite sides Opposite vertices Opposite sides Opposite vertices

n

2n

2


46. Translations via the vector (2, �3)

47. (a � 2, b � 3)48. Rotation 90�. Center of rotation: O with coordinates (0, 0).

49. (�b, a)50. Glide Translation: Line of reflection is the x-axis, translations is via the vector (�3,0)

51. (a � 3, �b)52.

A�

AO


53.

54.

55. (a)

(b) A�(0, 0), B�(0, 2) and C(1, 2)

(c)

56. (a)

(b) About 1.5

A

OB B�

A�

a13

a, 13

bb

Y

X

C

A B

C�

A�B�

P�

B�

C�A�

D�

O

A

B

C

D


57. Rotate �ABC 90� counterclockwise to �ABC. Then magnify �ABC using the size transformation of scale factor 4 and center the intersection of � and �.

58. Apply the size transformation with center (0, 0) and scale factor to �ABC. Follow this with a 180� rotation around (0, 0).

59. Reflections and and the 180� rotation around the intersection of the diagonals.60. The reflection with respect to .

61. 37.52 quart �

Proofs.62. Answers will vary. One approach might be to assign coordinates to the points: A(a, b), B(c, d),

P(n, m) and Q(r, t). Then the image of A, A�, under the first translation is A�(a � (r � n), b �(t � m)). Then the second translation moves A� to A so that A(a � (r � n) � (n � r), b �(t � m) � (m � t)) � A(a, b).

63. Answers will vary. A is a fixed point under both rotations. B rotates to B� so that �BAB� �125�. Then B� rotates to B so that �B�AB � 55�. Then �BAB � 180�. So B lies on the linecontaining .

64. Answers will vary.65. Answers will vary. One possible outline uses reflecting a triangle across on edge and showing

two angle pairs are congruent and therefore all three angles are congruent.

66. Answers will vary.

Applications67. Use a size transformation with scale factor 2.68. Rotate ABCD 90�, then 180�, then 270� to obtain the completed figure.69. Use a size transformation with scale factor 2.70. �ABC and �RST are both isosceles right triangles with leg lengths and (respectively)

and hypotenuse lengths and (respectively). (Note: and Thus the triangles are 45�–45� triangles with proportional

corresponding sides and congruent corresponding angles. This shows that �ABC � �RST . Wecan see by applying a size transformation with scale factor 2 and center (1, 1) to �ABC that �RSTis the image of �ABC under a size transformation with center (1, 1), scale factor 2 and translationvector (1, 3).

2252 + 2252 = 22102)252 + 252 = 21022110110

21515

C

BA

AB

1 gallon

4 quartsL 352 gallons

BE

ÍBD

!ÍAC

!

12

ÍBB

!ÍAA

!


71. Answers may very. Size transformation with center at the midpoint of and scale factor of 1/18. Reflection with respect to the perpendicular bisector of .

72. Size transformation with center at a corner and scale factor of 2 make the area 4 times as largeor a scale factor of 4 to make its length and width four times as long.

ABAB


Documents

GEOMETRY Chapter One—Answerstestbanktop.com/wp-content/uploads/2016/12/Downloadable... · 2016-12-31 · GEOMETRY Chapter Two—Answers True–False. 1. F:The triangles may be different